If the amplitude of sound is doubled and the frequency reduced to one- fourth the intensity of sound at the same point will

Options

A. incresase by a factor of 2
B. decrease by a factor of 2
C. decrease by a factor of 4
D. remains unchanged

Answer

To solve the problem, we need to understand how the intensity of sound is affected by changes in amplitude and frequency. The intensity (I) of sound is given by the formula:

\[ I \propto A^2 f^2 \]

where:
- \( I \) is the intensity,
- \( A \) is the amplitude,
- \( f \) is the frequency.

### Step-by-Step Solution:

1. **Identify the initial and final conditions**:
- Let the initial amplitude be \( A_1 \) and the initial frequency be \( f_1 \).
- According to the problem, the new amplitude \( A_2 = 2A_1 \) (doubled) and the new frequency \( f_2 = \frac{1}{4}f_1 \) (reduced to one-fourth).

2. **Write the expression for initial and final intensity**:
- The initial intensity \( I_1 \) can be expressed as:
\[ I_1 \propto A_1^2 f_1^2 \]
- The final intensity \( I_2 \) can be expressed as:
\[ I_2 \propto A_2^2 f_2^2 \]

3. **Substitute the new values into the intensity formula**:
- Substitute \( A_2 \) and \( f_2 \):
\[ I_2 \propto (2A_1)^2 \left(\frac{1}{4}f_1\right)^2 \]
- Simplifying this gives:
\[ I_2 \propto 4A_1^2 \cdot \frac{1}{16}f_1^2 \]
- This simplifies to:
\[ I_2 \propto \frac{4}{16} A_1^2 f_1^2 = \frac{1}{4} A_1^2 f_1^2 \]

4. **Relate the final intensity to the initial intensity**:
- Since \( I_1 \propto A_1^2 f_1^2 \), we can write:
\[ I_2 \propto \frac{1}{4} I_1 \]

5. **Conclusion**:
- This means that the final intensity \( I_2 \) is one-fourth of the initial intensity \( I_1 \). Therefore, the intensity of sound at the same point will decrease by a factor of 4.

### Final Answer:
The intensity of sound will decrease by a factor of 4.

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